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Using a 1-1 correspondence, we can show that \(p_n^* = p_n - p_{n - 1}\) (*) for all n >= 2.  So,

\(P^{*}(x) = 1 - x - x^3 + x^4 - x^5 + \dotsb\)

The degree 10 truncation of \(P(x) P^{*}(x)\) is then

\(1 + x^2 + x^4 - 2x^5 - x^6 + x^8 - 3x^9 + x^{10}\)

(*) "Theory of Partitions", George E. Andrews

There is something wrong with the question.....

It should be       ( (  16u² ) / (vx^-3  ) )^-1/2

then the answer would be

      v^1/2 /4 u x^3/2        B

10/x  =  x/6  ?

60 = x^2

x = sqrt 60     Is that right?

empy portion occupies a cylinder of  height 5

full part is a cylinder of 15        so 

  bottle is 1 part empty    3 parts full        1/4 of the bottle is empty     volume = 300 ml

If sec < 0 then    cos < 0         (sec is 1/cos)

sin^2 + cos^2 = 1

cos^2 = 1-sin^2

          = 1 - 1/25 =24/25

cos t =   -  sqrt 24  / 5                                 cot = cos / sin    I think you can do that one ...

yeah that's what i initially thought too, but the answer is actually 5 - sqrt(15)·i or  5 + sqrt(15)·i

The lateral surface area is 4(1/2*24*32).

4(1/2*24*32)

=4(12*32)

=1536

Notice that CD=US, UT=CB, and BD=ST. In other words, the two triangles are congruent.

Since CD=US:

3x+32=9x-4

6x=36

x=6

Since BD=ST:

y=2y-20

y=20

Hint,

any side must be shorter than the sum of the other two sides.

(Otherwise the sides could not join up)

Draw it and see what I am saying.

Now you can work it out.

To be congruent they have to be exactly alike.  Same shape, same size.  Let's look at same size. 

The hypotenuse of one has to equal the hypotenuse of the other.  So     10b  =  b + 36 

Solve for b                                                                                                   9b  =  36   

                                                                                                                      b  =  4  

Same with that leg                                                                                        9c  =  c + 16  

Solve for c                                                                                                    8c  =  16  

                                                                                                                       c  =  2  

_.

I actually need help with this problem too 

1. The numbers in which 3 occurs only once. This means that 3 is one of the digits and the remaining two digits will be any of the other 9 digits

You have 1*9*9 = 81 such numbers. However, 3 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 3 will appear only once.

2. The numbers in which 3 will appear twice. In these numbers, one of the digits is not 3 and it can be any of the 9 digits.

There will be 9 such numbers. However, this digit which is not 3 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 3 is written twice. Therefore, 3 is written 54 times.

3. The number in which 3 appears thrice - 333 - 1 number. 3 is written thrice in it.

Therefore, the total number of times the digit 3 is written between 1 and 999 is 243 + 54 + 3 = 300

Keep in mind squars are also rectangles. There are 7 squares.

rectangles composed of 2 squares: 8

rectangles composed of 3 squares: 2

rectangles composed of 4 squares(forms a larger square): 2

So in total I believe there are 7+8+2+2=15+4=19 rectangles.

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For the community: Please don't submit solutions to this problem. This is a homework problem for an online course that does not allow students to search for answers to specific problems outside of the school.

For the original poster: We realize that homework can be challenging. If you wish to receive help from the staff or other students, we encourage you to use the resources that the online classes provide. Please don't ask or search online for homework help. We understand that it's common in today's information age to look for resources online, and in some contexts, that's a great thing! However, it's against our Honor Code. You can ask for help on the message boards, and you can learn from the official solution after you submit your answer.

For the community: Please don't submit solutions to this problem. This is a homework problem for an online course that does not allow students to search for answers to specific problems outside of the school.

For the original poster: We realize that homework can be challenging. If you wish to receive help from the staff or other students, we encourage you to use the resources that the online classes provide. Please don't ask or search online for homework help. We understand that it's common in today's information age to look for resources online, and in some contexts, that's a great thing! However, it's against our Honor Code. You can ask for help on the message boards, and you can learn from the official solution after you submit your answer.

...? where I'll fix it

Similar triangles:  10 : x  =  x : 6

If  q(x)  =  5 - x² 

then:  q(r)  =  5 - r²            (substitue 'r' for 'x')

q(r) - 5  =  (5 - r²) - 5  =  -r² 

q(r) - q(5)  =  (5 - x²) - (5 - 5²)  =   (5 - x²) - (5 - 25​)  =  (5 - x²) - (-20​) =  5 - x² + 20  =  25 - x²  

Hint:  the denominator can be factored....then often things in the numerator and denominator 'cancel out'

the denominator can be factored  to  4 (6x^2-x-2)

                        which factors further to   4 (3x-2)(2x+1)          that should help ya !!